Nonuniform Acceleration
Ques 21: xcoordinate of a particle moving along this axis is: x = (2 + t^{2} + 2/3). Here, x is in metre and t in seconds. Find:
(a) position of particle from where it started its journey,
(b) initial velocity of particle and
(c) acceleration of particle at t = 2s.
Ans: (a) x = 2.0 m
(b) zero
(c) 26 ms^{2}^{}Sol:
Ques 22: The velocity of a particle moving in a straight line is decreasing at the rate of 3 m/s per metre of displacement at an instant when the velocity is 10 m/s. Determine the acceleration of the particle at this instant.
Ans:  30 ms^{2}
Sol:
Ques 23: The position o f a particle along a straight line is given by s = (t^{3}  9t^{2}  15t) m, here t is in seconds. Determine its maximum acceleration during the time interval 0 < / < 10s.
Ans: 42 ms^{2}
Sol: s = t^{3}  9t^{2}  15t
∴
ie,
Acceleration (a) in the interval 0 < t < 10 s will be maximum at t = 10s
a(at t = 10 s) = (6 x 10)  18
= 42 m/s^{2}
Ques 24: The acceleration of a particle is given by a (t) = (3.00 m/s^{2})  (2.00 m/s^{3})t.
(a) Find the initial speed v_{0} such that the particle will have the same xcoordinate at t = 5.00s as it had at t = 0.
(b) What will be the velocity at t = 5.00s?
Ans: (a) 0.833 ms^{1}
(b)  9.17 ms^{1}
Sol:
or ∫dv = ∫(3  2t) dt
or v = 3t  t^{2} + c
or v = 3t  t^{2} + v_{0} [as at t = 0, v = v_{0}]
or ∫ds = ∫ (3t  t^{2} + v_{0}) dt
or
(a) Displacement at = Displacement at
⇒ v_{0} = 5/6 = 0.833 s
(b) v = 3 t  t^{2} + v_{0}
Velocity at t (= 5.00 s)
= 3.5  5^{2} + 0.833
= 15  25 + 833
=  9.167 m/s
Ques 25: A particle moves along a horizontal path, such that its velocity is given by v = (3t^{2}  6t) m/s, where t is the time in seconds. If it is initially located at the origin O, determine the distance travelled by the particle in time interval from t = 0 to t = 3.5s and the particle’s average velocity and average speed druing the same time interval.
Ans: 14.125 m, 1.75 ms^{1}, 4 .0 3 ms^{1}
Sol: v = 3t^{2}  6t
∴ ∫ds = ∫(3t ^{2} 6t) dt
or s = t^{3}  3t^{2} + c
or s = t^{3}  3t^{2} = t^{2} (t  3)
= 3.5 x 0.5
= 1.75 m/s
(b) Distance covered
= 4.036 m/s
Ques 26: A particle travels in a straight line, such that for a short time 2s < t < 6s, its motion is described by v = (4/a) m/s, where a is in m/s^{2}. If v = 6 m/s when t = 2s, determine the particle’s acceleration when t = 3s.
Ans: 0.603 ms^{2}
Sol: v = 4/a
i.e., va = 4 …(i)
Now, at t = 2 s, v = 6 m/s
∴
Thus, C = 10.
i.e., …(ii)
∴ at t = 3 s,
Substituting above found value of v in Eq.
(i),
i.e.,
= 0.603 m/s^{2}
Ques 27: If the velocity v of a particle moving along a straight line decreases linearly with its displacement from 20 m/s to a value approaching zero at s = 30 m, determine the acceleration of the particle when s = 15 m.
Ans: ( 20/3) ms^{2}
Sol: According to question, the velocity of the particle varies as shown in figure.
∴
or 3v + 2s = 60
or 3 v = 60  2s …(i)
Differentiating above equation w.r.t. time t
or 3a =  2v
or
or
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